I was recently reading this question on the relation between Mordell's Theorem and FLT for the $n=4$ case. Reading more carefully Knapp's book, and more precisely the chapter where he discusses the relation described in the question, some questions have arisen:
With Mordell's Theorem in hand, we know that the passage $P\mapsto \frac12 P$ cannot go on forever.
I have been trying to understand this claim, i.e. that for any finitely generated abelian group $\Gamma$ the passage the passage $P\mapsto \frac12 P$ cannot go on forever. Thus, we have to prove, that there must exist some $n\in \mathbb{N}$ such that there is no $P'\in \Gamma$ satisfying $P = 2^nP'$. However, I cannot see how this should follow from $\Gamma$ being finitely generated. I have tried to use the Structure Theorem for finitely generated abelian groups, but with no success so far.
Later in that same page, Knapp writes
The hard step in the proof of Mordell's Theorem will be to prove that $E(\mathbb{Q})/2E(\mathbb{Q})$ is finite, so that the descent passage $P\mapsto \frac12 P$ can proceed for all but a finite number of situations.
Which makes me wonder if the claim doesn't follow alone from $\Gamma$ being finitely generated, but also from $\Gamma/2\Gamma$ being finite. (Are there finitely generated $\Gamma$ such that the descent passage $P\mapsto \frac12 P$ can go on forever?). Yet I don't understand the second quote either. Don't we want to prove that the descent passage cannot go on forever? Why then »for all but a finite number of situations«? And how does this follow from $\Gamma/2\Gamma$ being finite?
I know, depending on the perspective, this post might contain more than one question, but I felt that they can all be answered by the same and one answer. In particular, I believe that the key claim here is that $\Gamma/2\Gamma$ being finite implies that $P\mapsto \frac12 P$ cannot go on forever. Could you give me a hint on how to show that? Thanks in advance!
It's not true that $\Gamma$ being finitely generated implies that not every element of $\Gamma$ can be halved. (And it doesn't help to assume that $\Gamma/2\Gamma$ is finite, since that already follows from $\Gamma$ being finitely generated.) For instance, $\Gamma$ could be trivial, or more generally could be finite of odd order. So there must be some additional input that Knapp is glossing over in the statement you quote. If, for instance, you additionally know that your point $P$ is not a torsion point of the elliptic curve, then you can deduce that it cannot be divisible by $2^n$ for all $n$. This follows easily from the classification of finitely generated abelian groups: if $P$ is not torsion, then when you decompose $\Gamma$ as a direct sum of cyclic groups $P$ must have nonzero coordinate on some copy of $\mathbb{Z}$, and then $P$ cannot be divisible by $2$ any more times than that nonzero coordinate is.